Bzoj4289: PA2012 Tax

题面

传送门

Sol

巧妙的建图+\(Dijkstra\)
考虑把边看成点，那么显然暴力建图的边数是\(m^2\)的
考虑优化
把\(max(a, b)\)变成\(a+max(b-a,0)\)
把每个点连出的边按权值从小到大排序
每个边向后面的边连\(b-a\), 后面向前面连\(0\)
连向它的边向连出去的边连\(a\)
新建\(S\)点，向\(1\)连出的边连，新建\(T\)，连向\(n\)的边连\(T\)

没开long long WA一片

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e5 + 5);
typedef int Arr[_];
 
IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}
 
Arr first, vis;
ll dis[_];
int n, m, cnt, S, T;
struct Data{
    int u; ll dis;
 
    IL int operator <(RG Data B) const{
        return dis > B.dis;
    }
};
priority_queue <Data> Q;
struct Edge{
    int to, next, w;
} edge[_ << 2];
struct Link{
	int v, w, id;

	IL int operator <(RG Link B) const{
		return w < B.w;
	}
};
vector <Link> G[_];

IL void Add(RG int u, RG int v, RG int w){
    edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}
 
IL void Dijkstra(){
    Fill(dis, 63), dis[S] = 0, Q.push((Data){S, 0});
    while(!Q.empty()){
        RG Data x = Q.top(); Q.pop();
        if(vis[x.u]) continue;
        vis[x.u] = 1;
        for(RG int e = first[x.u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(dis[x.u] + w < dis[v]) Q.push((Data){v, dis[v] = dis[x.u] + w});
        }
    }
}
 
int main(RG int argc, RG char *argv[]){
	Fill(first, -1), n = Input(), m = Input();
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input(), v = Input(), w = Input();
		G[u].push_back((Link){v, w, i});
		G[v].push_back((Link){u, w, m + i});
	}
	T = m + m + 1; RG int tmp = m;
	for(RG int i = 1; i <= n; ++i){
		sort(G[i].begin(), G[i].end()); RG int l = G[i].size();
		for(RG int j = 0; j < l; ++j){
			Add((G[i][j].id > tmp ? G[i][j].id - m : G[i][j].id + m), G[i][j].id, G[i][j].w);
			if(i == 1) Add(S, G[i][j].id, G[i][j].w);
			if(G[i][j].v == n) Add(G[i][j].id, T, G[i][j].w);
		}
		for(RG int j = 0; j < l - 1; ++j){
			Add(G[i][j].id, G[i][j + 1].id, G[i][j + 1].w - G[i][j].w);
			Add(G[i][j + 1].id, G[i][j].id, 0);
		}
	}
    Dijkstra();
    printf("%lld\n", dis[T]);
    return 0;
}